TheoremStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-07

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

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Invariance of dimension for euclidean spaces

Statement

For nonnegative integers $m, n$ , a homeomorphism $R^{m} \to R^{n}$ implies $m = n$ .

Facts & Assumptions

Given: The objects and hypotheses in the statement above.

[F1]

For $n \geq 1$ , ${\tilde{H}}_{k} (S^{n}; G)$ is $G$ for $k = n$ and $0$ otherwise. For $S^{0}$ , ${\tilde{H}}_{0} (S^{0}; G) ≅ G$ and all other reduced groups vanish. Thus $H_{0} (S^{n}; G) ≅ G$ for $n \geq 1$ , whereas $H_{0} (S^{0}; G) ≅ G \oplus G$ . (Homology of spheres)

[F2]

If $f, g : X \to Y$ are homotopic continuous maps, then for every $n \geq 0$ and every abelian group $G$ the induced homomorphisms on singular homology agree: $H_{n} (f_{#}) = H_{n} (g_{#}) : H_{n}^{s i n g} (X; G) \to H_{n}^{s i n g} (Y; G) .$ (Homotopic maps induce the same map on singular homology)

Proof

1.1

The space $R^{0}$ is a singleton, whereas $R^{k}$ contains at least two points for $k > 0$ . Thus if one dimension is zero, a homeomorphism forces the other to be zero.

given

1.2

Suppose $m, n > 0$ . A homeomorphism restricts to $R^{m} ∖ {0} ≅ R^{n} ∖ {f (0)}$ . Translate the omitted target point to zero. For $k > 0$ the formula $R (x, t) = ((1 - t) + t / ∣ x ∣) x$ is a strong deformation retraction of $R^{k} ∖ {0}$ onto $S^{k - 1}$ : its scalar is positive and equals one when $∣ x ∣ = 1$ .

givenalgebra

2.1

Homotopy invariance in F2, restricted to the augmentation kernels in degree zero, now identifies the reduced homology of these spheres. By F1 with integral coefficients, the only nonzero reduced group of $S^{k - 1}$ is $Z$ in degree $k - 1$ , including $k = 1$ . Equality of this support forces $m - 1 = n - 1$ and hence $m = n$ .

F1F2step 1.2∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

9 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Miller, Algebraic Topology I lecture notes, Corollaries 10.5 and 10.6, pp.23–24 (standard reference, not scraped)